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Killing vecotrs of Schwarzschild metric

  1. Apr 1, 2014 #1
    I have a question, whether is possible to looking for Killing Vectors (KV) in this way (I know about general solution):[itex] [/itex]
    From Schwarzschild metric I can see two KV [itex]\frac{\partial}{\partial t} [/itex] and [itex]\frac{\partial}{\partial\phi} [/itex]. Then I see that other trivial KV arent there. Metric in dt and dr is independant on [itex]\theta, \phi [/itex] so I suppose I can "split" metric and looking for KV just in spherical part [itex]d\theta^2+\sin^2{\theta}d\phi^2 [/itex].
    Can I suppose transformation this metric to the form: [itex]d\alpha^2+d\beta^2 [/itex] and claim the [itex]\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta} [/itex] are KV?
  2. jcsd
  3. Apr 1, 2014 #2


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    No. You won't be able to find such a transformation.

    Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of ##so(3)##, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.
  4. Apr 1, 2014 #3
    Do I need know all transformation? I thought that differencial form of it could be enough. If I suppose [itex]\alpha=\alpha(\theta,\phi), \beta=\beta(\theta,\phi) [/itex], than I can write [itex]d\alpha=\alpha_{\theta}+\alpha_{\phi} [/itex], [itex]d\beta=\beta_{\theta}+\beta_{\phi} [/itex], where subscript means differentation. When I put it back to the [itex] ds^2=d\alpha^2+d\beta^2[/itex] I get 3 eq.:
    [itex]1=\alpha_{\theta}^2+\beta_{\theta}^2 [/itex]
    [itex]0=\alpha_{\theta}\alpha_{\phi}+\beta_{\theta}\beta_{\phi} [/itex]
    [itex]\sin^2{\theta}=\alpha_{\phi}^2+\beta_{\phi}^2 [/itex].
    From them (if from first guess [itex]\alpha_{\theta}=\cos{\phi} [/itex], [itex]\beta_{\theta}=\sin{\phi} [/itex]) I get [itex]d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi [/itex], [itex]d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi [/itex].
    Multiplying by [itex]g^{\mu\nu} [/itex] I hope can get [itex]\frac{\partial}{\partial\alpha} [/itex], [itex]\frac{\partial}{\partial\beta} [/itex]. What is wrong?
  5. Apr 1, 2014 #4


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    If you could put the metric in that form, it would be FLAT. It's not flat, it's a sphere.
  6. Apr 1, 2014 #5


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    This clearly doesn't reduce to the metric on the 2-sphere: simple substitution yields ##d\alpha^2 + d\beta^2 = d\theta ^2 + \sin^2 \theta d\phi^2 + (\sin \theta \cos\phi - \sin 2\phi )\sin\theta d\theta d\phi##.

    You simply won't be able to put the 2-sphere metric in the form ##ds^2 = d\alpha^2 + d\beta^2## and Bill has elucidated above why this is impossible.
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